---
tag: 86AN7M33VZ
kind: proof
name: Proof of the reverse triangle inequality
msc: [26A03]
proves: 0C9S13WKMF
uses: [03HF5PMY5T, 00S7C68D51]
sources:
  - url: https://proofwiki.org/wiki/Reverse_Triangle_Inequality/Real_and_Complex_Fields
---

## Proof

1. Let $a$ and $b$ be real numbers. For each real $t$, $\abs{-t} = \abs{t}$, by the definition of the [[absolute value|00S7C68D51]].
2. By the [[triangle inequality|03HF5PMY5T]], $\abs{a} = \abs{(a - b) + b} \le \abs{a - b} + \abs{b}$. So $\abs{a} - \abs{b} \le \abs{a - b}$.
3. Exchange $a$ and $b$: $\abs{b} - \abs{a} \le \abs{b - a} = \abs{a - b}$, by step 1.
4. Let $s = \abs{a} - \abs{b}$. By steps 2 and 3, $s \le \abs{a - b}$ and $-s \le \abs{a - b}$.
5. By definition, $\abs{s}$ equals $s$ or $-s$. In both cases, $\abs{s} \le \abs{a - b}$ by step 4.
6. So $\bigl| \abs{a} - \abs{b} \bigr| \le \abs{a - b}$.

## Idea

Write $a$ as $(a - b) + b$ and apply the triangle inequality. This bounds $\abs{a} - \abs{b}$, and the same argument with $a$ and $b$ exchanged bounds its negative.